What is the equation of the normal line of f(x)=6x^3-12x^2-x-1 at x=2?

Sep 7, 2016

$x + 23 y + 67 = 0$

Explanation:

$f \left(x\right) = 6 {x}^{3} - 12 {x}^{2} - x - 1$.

When $x = 2 , y = f \left(2\right) = 48 - 48 - 2 - 1 = - 3$.

Thus, normal to $f$ at #(2,-3) is reqd.

We know that, the slope of the tgt. at the pt.$\left(2 , - 3\right) \text{ is } f ' \left(2\right)$.

We have, $f ' \left(x\right) = 18 {x}^{2} - 24 x - 1 \Rightarrow f ' \left(2\right) = 72 - 48 - 1 = 23$.

Since, normal is $\bot$ to the tgt., its slope at $\left(2 , - 3\right)$ is $- \frac{1}{23}$.

Summing up the information about the normal, it has slope$- \frac{1}{23} ,$

&, passes thro. $\left(2 , - 3\right)$.

Using, Point-Slope Form for the Normal, its. eqn. is given by,

$y - \left(- 3\right) = - \frac{1}{23} \left(x - 2\right) , \mathmr{and} , 23 \left(y + 3\right) + \left(x - 2\right) = 0$;

i.e., $x + 23 y + 67 = 0$

Enjoy Maths.!